Question

How many mL of 0.200 M NaOH must be added to 31.35 mL of a 0.0500 F solution of fumaric acid (trans-butenedioic acid) to make a buffer of pH 4.50?

Answer 1

pH of acidic buffer = pka + log(base(or) salt/acid)

pka of fumaric acid = 3.55

pH = 4.5

no of mol of fumaric acid = 31.35*0.05 = 1.5675 mol

no of mol of NaOH = x mol

4.5 = 3.55+log(x/(1.5675-x))

x = 1.41

volume of NaOH required = n/M

                         = 1.41/0.2

                         = 7.05 ml